sum_(n=1)^oo n^k*e^(-n) Use the Integral Test to determine the convergence or divergence of the series, where k is a positive integer.

The Integral test says that this sum will converge if and only if this integral also converges:int_{1}^{infty} x^k e^{-x} dx When integrating this, we would use integration by parts, and we would need to use it k times. The first set of parts is u = x^k, dv = e^{-x} dx, du = k x^{k-1} dx, v = -e^{-x}
 
int u dv = u v - int v du = - x^k e^{-x}|_1^infty + int_{1}^{infty} k x^{k-1} e^{-x} dx
Then we repeat for u_1 = x^{k-1} , and so on until we have only the e^{-x} term left.
But the important thing is that the last term would only be in terms of a constant times int e^-x dx , which clearly converges; and then all the other terms would look like this, for some integer 1 leq p leq k and some constant C:
C x^p e^{-x} |_{1}^{infty} The value of this term at x = 1 we can simply calculate; no problem there, it will be some finite number. The limit as x goes to infinity we can also determine by the fact that e^x always increases faster than any polynomial as x gets very large, and thus for any value of p, this limit must be zero.Thus, we have k-1 terms that are finite (zero minus a finite value), plus one final term that is a convergent integral. Therefore the whole integral converges; therefore the sum converges.